Double Integrals 6 Let's evaluate the double integrals with y=x^2 as one of the boundaries.
Double Integrals 6
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- Welcome back.
- In the last video we were just figuring out the volume under
- the surface, and we had set up these integral bounds.
- So let's see how to evaluate it now.
- And look at this.
- I actually realized that I can scroll things, which is quite
- useful because now I have a lot more board space.
- So how do we evaluate this integral?
- Well, the first integral I'm integrating with respect to x.
- I'm adding up the little x sums.
- So I'm forming this rectangle right here.
- Or you could kind of view it I'm holding y constant and
- integrating along the x-axis.
- I should switch colors.
- So what's the antiderivative of x y squared with respect to x?
- Well it's just x squared over 2.
- And then I have the y squared-- that's just a
- constant-- all over 2.
- And I'm going to evaluate that from x is equal to 1 to x is
- equal to the square root of y, which you might be daunted by.
- But you'll see that it's actually not that bad
- once you evaluate them.
- And then let me draw the outside of the integral.
- This is y is equal to 0 to y is equal 1.
- Now, if x is equal to 1 this expression becomes
- y squared over 2.
- y squared over 2, minus-- now if x is equal to square root
- of y, what does this expression become?
- If x is equal to the square root of y, then
- x squared is just y.
- And then y times y squared is y to the third.
- So it's y to the third over 3.
- Fair enough.
- And now I take the integral with respect to y.
- So now I sum up all of these rectangles in the y direction.
- 0, 1.
- This is with respect to y.
- And that's cool, right?
- Because when you take the first integral with respect to x you
- end up with a function of y anyway, so you might as well
- have your bounds as functions of y's.
- It really doesn't make it any more difficult.
- But anyway, back to the problem.
- What is the antiderivative of y squared over 2 minus
- y to the third over 3?
- Well the antiderivative of y squared-- and you have to
- divide by 3, so it's y cubed over 6.
- Minus y to the fourth-- you have to divide by 4.
- Minus y to the fourth over-- did I mess up some place?
- No, I think this is correct. y to the fourth over 12.
- Oh wait.
- How did I get a 3 here?
- That's where I messed up.
- This is a 2, right?
- Let's see.
- x is square root of y.
- Yeah, this is a 2.
- I don't know how I ended up.
- Square root of y squared is y, times y squared
- y to the third over 2.
- And then when I take the integral of this
- it's 4 times 2.
- Got to make sure I don't make those careless mistakes.
- That's the tough part.
- I just want to make sure that you got that too.
- I hate it when I do that.
- But I don't want to re-record the whole video.
- So when I evaluated this-- right.
- This is right, and then I take the antiderivative of y to the
- third over 2, I get y to the fourth over 8.
- And now I evaluate this at 1 and 0.
- And that give us what?
- 1/6 minus 1/8 minus-- well both of these when you evaluate
- them, are going to be 0's.
- So this is going to be another 0 minus 0.
- So you don't have to worry about that.
- So what's 1/6 minus 1/8?
- Let's see.
- That's 4 minus 3 over 24, which is equal to 1/24-- is the
- volume of our figure.
- So this time, the way we just did it, we took the integral
- with respect to x first, and then we did it with
- respect to y.
- Let's do it the other way around.
- So let me erase some things.
- And hopefully I won't make these careless mistakes again.
- I'll keep this figure, but I'll even erase this one.
- Let me erase all of this stuff.
- We have room to work with.
- So I kept that figure.
- But let me redraw just the xy plane, just so we get
- the visualization right.
- It's more important to visualize the xy plane in
- these problems than it is to visualize the
- whole thing in 3-d.
- That's the y-axis, that's the x-axis.
- Our upper bound, you can say, is the graph y is
- equal to x squared.
- Or, you could view it as the bound x is equal to
- the square root of y.
- That is x is equal to 1, that's y is equal to 1.
- And we care about the volume above the shaded region.
- That shaded region is this yellow region right here.
- And let's draw our da.
- I'll draw a little square actually.
- And I'll do it in magenta.
- So that's our little da.
- And the height is dy.
- That's a y.
- And it's with this dx.
- So the volume above this little square-- that's the same
- thing as this little square.
- Just like we said before, the volume above it is equal to
- the value of the function.
- The height is the value of the function, which is x y squared.
- And then we multiply it times the area of the base.
- Well the area of the base, you could say it's da, but we know
- it's really dy times dx.
- I didn't have to write that times there.
- You can ignore it.
- And I wrote the y first just because we're
- going to integrate with respect to y first.
- We're going to sum in the y direction.
- So what does summing in the y direction mean?
- It means we're going to add that square to that
- square to that square--
- Excuse me.
- So we're adding all the dy's together, right?
- So my question to you is, what is the upper bound on the y?
- Well once again, we bump into the curve.
- So the curve is the upper bound when we go upwards.
- And what is the upper bound on the curve?
- Well we're holding x fixed, so for any given
- x what is this point?
- Well that's going to be x squared.
- Because this is the graph of y equals x squared.
- So our upper bound is y is equal to x squared.
- And what's our lower bound?
- We can keep adding these squares down here.
- We're adding all the little changes in y.
- So what's our lower bound?
- Well our lower bound is just 0.
- That was pretty straightforward.
- So this expression, as it is written right now, is the
- volume above this rectangle.
- Let me draw it.
- Is the volume above this rectangle.
- These are the same rectangles.
- The volume above that rectangle.
- Now what we want to do is add up all the dx's together, and
- we'll get the volume above the entire surface.
- So that rectangle, now we'll add it to another
- dx one, dx one like that.
- So what's the upper and lower bounds on the x's?
- We're going from x is equal to 0, right?
- We don't bump into the graph when we go all the way down.
- So we go from x is equal to 0.
- And then our upper bound is x is equal to 1.
- So x is equal to 0, x is equal to 1.
- And in general, one way to think about it is when you're
- doing kind of the last sum or the last integral, you really
- shouldn't have variable boundaries at this point.
- Because our final answer has to be a number, assuming
- that we're not dealing with something very, very abstract.
- But our final answer is going to have a number.
- So if you had some variables here you probably
- did something wrong.
- And it's always useful, I think, to draw that little da
- and then figure out-- OK, I'm summing to dy first.
- When I go upwards I bump into the curve.
- What is that upper bound if x is constant?
- Oh, it's x squared. y is equal to x squared.
- If I go down I bump into the x-axis, or y is equal to 0.
- And so forth and so on.
- So now let's just evaluate this and confirm that
- we get the same answer.
- So we're integrating with respect to y first.
- So that's x y to the third, over three.
- At x squared and 0.
- And then we have our outer integral. x goes from 0 to 1.
- If we substitute x squared in for y.
- x squared to the third power is x to the sixth.
- x to the sixth times x-- let me write that.
- So we have x times x squared to the third power, over 3.
- That equals x to the seventh.
- Right? x squared to the third-- you multiply the exponents, and
- then you add these. x to the seventh over 3, minus this
- evaluated with y is 0.
- But that's just going to be 0.
- And then we evaluate that from 0 to 1, dx.
- We're almost there.
- Increment the exponent.
- You get x to the eighth over 8.
- And we already have a 3 down there, so it's over 24.
- And you evaluate that from 0 to 1.
- And I think we get the same answer.
- When you evaluate it at 1 you get 1/24 minus 0.
- So once again, when we integrate in the other order
- you still get the volume of the figure, being 1/4,
- whatever, cubic units.
- Anyway, see you in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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